Chapter 4, Problem 50E

Interpretation Introduction

Interpretation:

The chemical equations are to be balanced for the given reactions.

Concept introduction:

A chemical equation is a symbolic depiction of a chemical reaction that contains symbols and formula.

In a chemical reaction, the starting substances on the left side are known as reactants, while the substances on the right side are known as products.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 50E

Solution: The balanced chemical equation is given as follows:

$3\text{PbO}+2{\text{NH}}_3\to 3\text{Pb}+{\text{N}}_2+3{\text{H}}_2\text{O}$

${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}+6{\text{H}}_2\text{O}\to 3\text{Mg}{\left(\text{OH}\right)}_2+2{\text{NH}}_3$

${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

a) $\text{PbO}+{\text{NH}}_3\to \text{Pb}+{\text{N}}_2+{\text{H}}_2\text{O}$

$\text{H}$ atom is balanced by multiplying the coefficient $\text{2}$ with ${\text{NH}}_{\text{3}}$ and the coefficient $3$ with ${\text{H}}_{\text{2}}\text{O}$ as follows:

$\text{PbO}+2{\text{NH}}_3\to \text{Pb}+{\text{N}}_2+3{\text{H}}_2\text{O}$

In the next step, $\text{O}$ atom is balanced by multiplying the coefficient $\text{3}$ with $\text{PbO}$ and further, $\text{Pb}$ atom is balanced by multiplying the coefficient $\text{3}$ with $\text{Pb}$ as follows:

$\text{3}\text{PbO}+2{\text{NH}}_3\to 3\text{Pb}+{\text{N}}_2+3{\text{H}}_2\text{O}$

b) ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}+{\text{H}}_2\text{O}\to \text{Mg}{\left(\text{OH}\right)}_2+{\text{NH}}_3$

$\text{Mg}$ atom is balanced by multiplying the coefficient $3$ with $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ as follows:

${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}+{\text{H}}_2\text{O}\to 3\text{Mg}{\left(\text{OH}\right)}_2+{\text{NH}}_3$

In the next step, $\text{O}$ atom is balanced by multiplying the coefficient $6$ with ${\text{H}}_{\text{2}}\text{O}$ as follows:

${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}+6{\text{H}}_2\text{O}\to 3\text{Mg}{\left(\text{OH}\right)}_2+{\text{NH}}_3$

Now, in the final step, $\text{H}$ atom is balanced by multiplying the coefficient $2$ with ${\text{NH}}_3$ as follows:

${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}+6{\text{H}}_2\text{O}\to 3\text{Mg}{\left(\text{OH}\right)}_2+2{\text{NH}}_3$

c) ${\text{C}}_3{\text{H}}_8+{\text{O}}_2\to {\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

$\text{H}$ atom is balanced by multiplying the coefficient $4$ with ${\text{H}}_2\text{O}$ as follows:

${\text{C}}_3{\text{H}}_8+{\text{O}}_2\to {\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

In the next step, $\text{C}$ atom is balanced by multiplying the coefficient $\text{3}$ with ${\text{CO}}_2$ as follows:

${\text{C}}_3{\text{H}}_8+{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$

Now, in the final step, $\text{O}$ atom is balanced by multiplying the coefficient $5$ with ${\text{O}}_2$ as follows:

${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_{\text{2}}\text{O}$